Question

a)Consider the funtion f:R\{−1}→R defined by f(x)=2x+1 x+1. i) ii) iii) iv) Check that f(x) is well defined and 1−1. Check that f(x)=2 for any x∈R. Check that g:R\{2}→R given by g(x)=x−1 Further,check that g(x)=−1 for any x∈R. 2−x is well defined and 1−1. (20) (3) (2) (4) Check that (f◦g)(x)=x for x∈R\{2} and (g◦f)(x)=x for x∈R\{−1}.(4) b)Find the direction cosines of the perpendicular from the origin to the plane r·(6i+4j+2√ 3k)+2=0.

Answer 1

Answer on Question #52685, Math, Linear Algebra

a) Consider the function $f: \mathbb{R} \setminus \{-1\} \to \mathbb{R}$ defined by $f(x) = 2x + 1 / x + 1$. i) ii) iii) iv) Check that $f(x)$ is well defined and 1-1. Check that $f(x) = 2$ for any $x \in \mathbb{R}$. Check that $g: \mathbb{R} \setminus \{2\} \to \mathbb{R}$ given by $g(x) = x - 1$. Further, check that $g(x) = -1$ for any $x \in \mathbb{R}$. 2-x is well defined and 1-1. (20) (3) (2) (4) Check that $(f * g)(x) = x$ for $x \in \mathbb{R} \setminus \{2\}$ and $(g * f)(x) = x$ for $x \in \mathbb{R} \setminus \{-1\}$. (4) b) Find the direction cosines of the perpendicular from the origin to the plane $r \cdot (6i + 4j + 2\sqrt{3}k) + 2 = 0$.

Answer:

a) $f: \mathbb{R} \setminus \{-1\} \to \mathbb{R}$ $f(x) = \frac{2x + 1}{x + 1}$

$$f(x_1) = f(x_2) \Rightarrow \frac{2x_1 + 1}{x_1 + 1} = \frac{2x_2 + 1}{x_2 + 1} \Rightarrow (2x_1 + 1)(x_2 + 1) = (2x_2 + 1)(x_1 + 1)$$ $$2x_1x_2 + 2x_1 + x_2 + 1 = 2x_1x_2 + x_1 + 2x_2 + 1 \Rightarrow x_1 = x_2$$ $$f(x) = \frac{2x + 1}{x + 1} = 2 \Rightarrow 2x + 1 = 2x + 2 \Rightarrow 1 \neq 2$$ $$x \notin \mathbb{R}$$ $$f(x) = \frac{2x + 1}{x + 1} = y \Rightarrow 2x + 1 = y(x + 1) \Rightarrow (2 - y)x = y - 1 \Rightarrow x = \frac{y - 1}{2 - y}$$ $$\mathbb{R} \setminus \{2\} \to \mathbb{R}: g(x) = x - 1$$ $$g(x_1) = g(x_2) \Rightarrow x_1 - 1 = x_2 - 1 \Rightarrow x_1 = x_2$$ $$g(x) = -x + 1 = -1 \Rightarrow x = 2$$ $$\text{but } \mathbb{R} \setminus \{2\} \to \mathbb{R}$$ $$(g * f)(x) = g((f(x))) = g\left(\frac{2x + 1}{x + 1}\right) = -\frac{2x + 1}{x + 1} + 1 = \frac{-2x + 1 + (x + 1)}{x + 1} = -\frac{x}{x + 1}$$ $$x \neq -1$$ $$(f * g)(x) = f((g(x))) = f(x - 1) = \frac{2(1 - x) + 1}{(1 - x) + 1} = \frac{3 - 2x}{2 - x}$$ $$x \neq 2$$

b) $r \cdot (6i + 4j + 2\sqrt{3}k) + 2 = 0$

The given equation can be written as

$$\vec{r} \cdot \left(6 \vec{i} + 4 \vec{j} + 2 \sqrt{3} \vec{k}\right) = -2$$ $$\left| \left(6 \vec{i} + 4 \vec{j} + 2 \sqrt{3} \vec{k}\right) \right| = \sqrt{6^{2} + 4^{2} + \left(2 \sqrt{3}\right)^{2}} = 8$$

Then $\frac{\vec{r} \cdot (6\vec{i} + 4\vec{j} + 2\sqrt{3}\vec{k})}{8} = -\frac{2}{8} \Rightarrow \vec{r} \cdot \left(\frac{3}{4}\vec{i} + \frac{1}{2}\vec{j} + \frac{\sqrt{3}}{4}\vec{k}\right) = -\frac{1}{4}$

which is the equation of the plane in the form $\vec{r} \cdot \vec{n} = d$

The direction cosines of $\vec{n}$ are $\frac{3}{4}, \frac{1}{2}, \frac{\sqrt{3}}{4}$.

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